<u>Answer:</u> The limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
- <u>For potassium sulfate:</u>
Molarity of potassium sulfate solution = 1.76 M
Volume of solution = 19.9 mL
Putting values in equation 1, we get:
- <u>For barium nitrate:</u>
Molarity of barium nitrate solution = 0.896 M
Volume of solution = 14.4 mL
Putting values in equation 1, we get:
The chemical equation for the reaction of potassium sulfate and barium nitrate follows:
By Stoichiometry of the reaction:
1 mole of barium nitrate reacts with 1 mole of potassium sulfate
So, 0.013 moles of barium nitrate will react with = of potassium sulfate
As, given amount of potassium sulfate is more than the required amount. So, it is considered as an excess reagent.
Thus, barium nitrate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 moles of barium nitrate produces 1 mole of barium sulfate.
So, 0.013 moles of barium nitrate will produce = of barium sulfate.
Now, calculating the mass of barium sulfate by using the equation:
- To calculate the number of moles, we use the equation:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.013 moles
Putting values in above equation, we get:
- To calculate the percentage yield of barium sulfate, we use the equation:
Experimental yield of barium sulfate = 2.49 g
Theoretical yield of barium sulfate = 3.03 g
Putting values in above equation, we get:
Hence, the limiting reagent is barium nitrate, the theoretical yield of barium sulfate is 3.03 grams and the percent yield of the reaction is 82.18 %.